Timeline for Does there exist a Haken manifold where all its incompressible surfaces are non-separating?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2018 at 0:40 | vote | accept | Anubhav Mukherjee | ||
| Nov 8, 2017 at 10:04 | comment | added | Anubhav Mukherjee | @ian agol yes, I was a bit sloppy while writing the first line. As you mentioned in the last line of your comment, I want to restrict it into connected components. | |
| Nov 8, 2017 at 4:10 | comment | added | Ian Agol | The first two sentences of your question are not quite correct. If a non-zero element of $H_2(M;\mathbb{Z})$ which is not primitive is represented by an embedded incompressible surface, then the surface must be disconnected, consisting of homologically parallel surfaces, hence separating. So of course in your question you want to restrict to connected incompressible surfaces. | |
| Nov 7, 2017 at 21:14 | comment | added | Ryan Budney | Yes, it's a semi-direct product of $\mathbb Z$ with $\mathbb Z^2$. | |
| Nov 7, 2017 at 19:56 | answer | added | Allen Hatcher | timeline score: 12 | |
| Nov 2, 2017 at 14:25 | comment | added | Anubhav Mukherjee | @RyanBudney whats it's fundamental group? Is it non-abelian? | |
| Nov 1, 2017 at 0:52 | comment | added | Ryan Budney | Here is a revised version of my previous comment. The specific example I gave did not work out, as there's a separating incompressible torus (boundary of a neighbourhood of the incompressible klein bottle). But if you take the torus bundle over the circle where the torus is "square" and the monodromy is order 4 then this manifold does appear to work as I believe the only incompressible surfaces are horizontal, i.e, the fiber. | |
| Oct 31, 2017 at 22:21 | comment | added | Ryan Budney | For the first part of Q2, plenty of Bieberbach manifolds should work, no? Say the orientable Bieberbach (euclidean) manifold that is a circle bundle over the Klein bottle. | |
| Oct 31, 2017 at 18:04 | answer | added | Sam Nead | timeline score: 3 | |
| Oct 31, 2017 at 17:40 | comment | added | Sam Nead | The three-torus is indeed an answer to your first question. It is closed, Haken, and all of its incompressible surfaces are two-tori, and are non-separating. Your second question is more difficult. | |
| S Oct 31, 2017 at 13:42 | history | suggested | Glorfindel | CC BY-SA 3.0 |
grammar corrections
|
| Oct 31, 2017 at 12:54 | review | Suggested edits | |||
| S Oct 31, 2017 at 13:42 | |||||
| Oct 31, 2017 at 12:47 | history | asked | Anubhav Mukherjee | CC BY-SA 3.0 |